Tag Archives: 基础、联系、序列和功能代写

基础、联系、序列和功能|MA10274 Foundations, connections, sequences and functions代写

0
Posted on by

Aims for foundations and connections: To improve the ability of the students to read and write mathematics well; to enable them to develop effective study skills for individual and group working; to introduce students to the beauty, the ubiquity and the importance of university-level mathematics.

这是一份Bath巴斯大学MA10265R作业代写的成功案

基础、联系、序列和功能|MA10274 Foundations, connections, sequences and functions代写

The second inequality is of the form $t^{2}>A$ and, as in Example 6 , a numerical illustration will help. If we have $t^{2}>4$ then $t$ could be positive or negative, giving two possibilities, namely $t>2$ or $t<-2$. So Case 2 becomes $x<0$ and $\left(x<\frac{3}{2}-\frac{\sqrt{13}}{2}\right.$ or $\left.x>\frac{3}{2}+\frac{\sqrt{13}}{2}\right) .$
The fact that $x$ is negative means that the third inequality cannot be satisfied, and so this case reduces to
$$
x<\frac{3}{2}-\frac{\sqrt{13}}{2} .
$$
Finally, putting together Cases 1 and 2 gives the complete solution as
$$
x<\frac{3}{2}-\frac{\sqrt{13}}{2} \text { or } 0<x<\frac{3}{2}+\frac{\sqrt{13}}{2} \text {. }
$$

英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

MA10274 COURSE NOTES ：

From the definition of the modulus we can calculate that $\left|z_{1}\right|=2$ and $\left|z_{2}\right|=\sqrt{2}$.
From the diagram we can find the angles, so $\theta_{1}=120^{\circ}=2 \pi / 3$ and $\theta_{2}=135^{\circ}=3 \pi / 4$. This gives all the information we need, and so
$$
z_{1}=2\left(\cos \frac{2 \pi}{3}+\mathrm{i} \sin \frac{2 \pi}{3}\right), \quad z_{2}=\sqrt{2}\left(\cos \frac{3 \pi}{4}+\mathrm{i} \sin \frac{3 \pi}{4}\right) .
$$
If we now use the polar product formula we obtain
$$
z_{1} z_{2}=2 \sqrt{2}\left(\cos \frac{17 \pi}{12}+\mathrm{i} \sin \frac{17 \pi}{12}\right) .
$$



基础、联系、序列和功能 Foundations, connections, sequences and functions MA10274

0
Posted on by

这是一份BATH巴斯大学MA10274作业代写的成功案例

基础、联系、序列和功能 Foundations, connections, sequences and functions MA10274
问题 1.

First we show that $|y|$ is necessarily 1 . In fact, if we let $c=(1-|y|) / 2$. then $0 \leq c \leq 1$. Considering the vectors
$$
x_{1}=\frac{(1-c) x+c y}{|y|} \text { and } y_{1}=\frac{(1-c)(y)+c x}{|y|}
$$
we find that $x_{1}, y_{1} \in B_{X}$ and $\left|x_{1}-y_{1}\right|=\varepsilon$. Therefore, .

证明 .

Rut
$$
\left|x_{1}+y_{1}\right|=\frac{1}{|y|}|\dot{x}+y|_{.} .
$$
Since $|y| \leq 1$, it follows from this last inequality and our choice of $x, y$ that il. $\left.y\right|^{\prime}=1$.

Having ascertained that $|y|=1$, what about $x$ ? Of course, if $|x|=1$, too. then we are done. Suppose $|x|<1$. Pick $\varphi \in S_{X}$ * so that
$$
\varphi\left(\frac{x+y}{|x+y|}\right)=1 \text {. }
$$

英国论文代写Viking Essay为您提供作业代写代考服务

MA10274 COURSE NOTES ：

$$
\left(2\left[1-\delta_{\left|x_{a} \ln z\right|}(k)\right]\right)^{p}<2 $$ Since given $x, y \in S_{X}$ for which $|x-y| \geq k$ we have $$ |x+y| \leq 2\left(1-\delta_{\left\{x_{n} \ln _{A 21}\right.}(k)\right) . $$ it follows that the continuous functions $\varphi(t)$ and $\chi(t)$ given by $$ \varphi(t)=|x+t y|^{p}, \quad \chi(t)=1+t^{p} $$ satisfy $\varphi(1)<\chi(1)$; consequently, there is an $\eta>0$ so that
$$
|x+t y|^{p} \leq 1+t^{p},
$$
whenever $|1-t| \leq \eta$. Of course, we can assume $\eta$ is very small, say $\eta<1$.
Claim. For any finitely nonzero sequence $\left(a_{m}\right)$ of scalars we have
$$
\frac{\eta}{2}\left|\sum_{m} a_{m} x_{m}\right| \leq\left|\left(a_{m}\right)\right|_{p} .
$$